Question

The solution of the differential equation $\left(1 - x^{2}\right)\frac{d y}{d x}-xy=1$ is (where, $\left|x\right| < 1,x\in R$ and $C$ is an arbitrary constant)

• A. $y\left(1 - x^{2}\right)=tan^{- 1}x+C$
• B. $y\sqrt{1 - x^{2}}=tan^{- 1}x+C$
• C. $y\sqrt{1 - x^{2}}=sin^{- 1}\left(x\right)+C$
• D. $y\cdot \left(1 - x^{2}\right)=sin^{- 1}x+C$

Answer 1

Answer: (C)

$\frac{d y}{d x}-\frac{x}{1 - x^{2}}y=\frac{1}{1 - x^{2}}$
$Ι.F.=e^{- \displaystyle \int \frac{x}{1 - x^{2}} d x}$
$=e^{\frac{1}{2} \displaystyle \int - \frac{2 x}{1 - x^{2}} d x}=e^{\frac{1}{2} ln \left(1 - x^{2}\right)}=\sqrt{1 - x^{2}}$
Hence, the solution of the differential equation is
$y\sqrt{1 - x^{2}}=\displaystyle \int \frac{\sqrt{1 - x^{2}}}{1 - x^{2}} d x$
$y\sqrt{1 - x^{2}}=\displaystyle \int \frac{1}{\sqrt{1 - x^{2}}} d x$
$y\sqrt{1 - x^{2}}=sin^{- 1}\left(x\right)+C$